JEE Main 2021 — Mathematics Calculus

If, $\beta =\underset{x\rightarrow 0}{\mathrm{lim}}{(\mathrm{cos}x)}^{\mathrm{cot}x}$β=x→0lim​(cosx)cotx (${1}^{\infty }$1∞form)

$$\beta ={e}^{\underset{x\rightarrow 0}{\mathrm{lim}}\mathrm{cot}x[\mathrm{cos}x-1]}$$β=ex→0lim​cotx[cosx−1]

$$\beta ={e}^{\underset{x\rightarrow 0}{\mathrm{lim}}\frac{-\mathrm{cos}x}{\mathrm{sin}x}[1-1+2{\mathrm{sin}}^{2}\frac{x}{2}]}$$β=ex→0lim​sinx−cosx​[1−1+2sin22x​]

$$\beta ={e}^{\underset{x\rightarrow 0}{\mathrm{lim}}\frac{-\mathrm{cos}x}{2\mathrm{sin}\frac{x}{2}.\mathrm{cos}\frac{x}{2}}[1-1+2{\mathrm{sin}}^{2}\frac{x}{2}]}$$β=ex→0lim​2sin2x​.cos2x​−cosx​[1−1+2sin22x​]

$$\beta ={e}^{\underset{x\rightarrow 0}{\mathrm{lim}}\frac{-\mathrm{cos}x.\mathrm{sin}\frac{x}{2}}{\mathrm{cos}\frac{x}{2}}}$$β=ex→0lim​cos2x​−cosx.sin2x​​

$$\beta ={e}^{0}$$β=e0

$$\beta =1$$β=1

$$\alpha =\underset{x\rightarrow \pi /4}{\mathrm{lim}}\frac{{\mathrm{tan}}^{3}x-\mathrm{tan}x}{\mathrm{cos}(x+\frac{\pi }{4})}$$α=x→π/4lim​cos(x+4π​)tan3x−tanx​

$$=\underset{x\rightarrow \pi /4}{\mathrm{lim}}\frac{\mathrm{tan}x(\mathrm{tan}x+1)(\mathrm{tan}x-1)}{\mathrm{cos}(x+\frac{\pi }{4})}$$=x→π/4lim​cos(x+4π​)tanx(tanx+1)(tanx−1)​

$$\alpha =\underset{x\rightarrow \pi /4}{\mathrm{lim}}\mathrm{tan}x\times \underset{x\rightarrow \pi /4}{\mathrm{lim}}(\mathrm{tan}x+1)\times \underset{x\rightarrow \pi /4}{\mathrm{lim}}\frac{(\mathrm{tan}x-1)}{\mathrm{cos}(x+\frac{\pi }{4})}$$α=x→π/4lim​tanx×x→π/4lim​(tanx+1)×x→π/4lim​cos(x+4π​)(tanx−1)​

$$\alpha =1\times (1+1)\times \underset{x\rightarrow \pi /4}{\mathrm{lim}}\frac{(\mathrm{tan}x-1)}{\mathrm{cos}(x+\frac{\pi }{4})}$$α=1×(1+1)×x→π/4lim​cos(x+4π​)(tanx−1)​

$$=2\underset{x\rightarrow \pi /4}{\mathrm{lim}}\frac{\mathrm{tan}x-1}{\mathrm{cos}(x+\frac{\pi }{4})}$$=2x→π/4lim​cos(x+4π​)tanx−1​

$$=2\underset{x\rightarrow \pi /4}{\mathrm{lim}}\frac{\mathrm{sin}x-\mathrm{cos}x}{\mathrm{cos}x.\mathrm{cos}(x+\frac{\pi }{4})}$$=2x→π/4lim​cosx.cos(x+4π​)sinx−cosx​

$$=-2\sqrt{2}\underset{x\rightarrow \pi /4}{\mathrm{lim}}\frac{(\frac{1}{\sqrt{2}}\mathrm{cos}x-\frac{1}{\sqrt{2}}\mathrm{sin}x)}{\mathrm{cos}x.\mathrm{cos}(x+\frac{\pi }{4})}$$=−22​x→π/4lim​cosx.cos(x+4π​)(2​1​cosx−2​1​sinx)​

$$\alpha =-2\sqrt{2}\underset{x\rightarrow \pi /4}{\mathrm{lim}}\frac{\mathrm{cos}(x+\frac{\pi }{4})}{\mathrm{cos}x.\mathrm{cos}(x+\frac{\pi }{4})}$$α=−22​x→π/4lim​cosx.cos(x+4π​)cos(x+4π​)​

$$\alpha =-4$$α=−4

So, the equation whose roots are $\alpha$α and $\beta$β is

$${x}^{2}+3x-4=0$$x2+3x−4=0

Hence, $a=1,b=3$a=1,b=3